我试图在包含'ls -l'输出的文件中使用完整路径grep文件名,但是无法正确匹配它。
执行字符串搜索的shell脚本中的行
pcline=`grep -w "$file1" $file2` # grep for file1 in file2 contents
如果我回显命令,命令的输出如下所示
grep -w run /home/rajesh/rootfs.layout
Expected
lrwxrwxrwx 1 root root 3 Aug 28 run
Actual
lrwxrwxrwx 1 root root 7 Aug 28 bin/run-parts
lrwxrwxrwx 1 root root 3 Aug 28 run
-rwxr-xr-x 1 root root 303 Aug 28 tests/aes/run.sh
-rwxr-xr-x 1 root root 445 Aug 28 tests/auto_ui/run.sh
-rwxr-xr-x 1 root root 320 Aug 28 tests/available_memory/run.sh
-rwxr-xr-x 1 root root 308 Aug 28 tests/fonts/run.sh
-rwxr-xr-x 1 root root 309 Aug 28 tests/html_config_page/run.sh
-rwxr-xr-x 1 root root 361 Aug 28 tests/ipc/run.sh
-rwxr-xr-x 1 root root 304 Aug 28 tests/JSON/run.sh
-rwxr-xr-x 1 root root 303 Aug 28 tests/log4cplus_cpp/run.sh
-rwxr-xr-x 1 root root 301 Aug 28 tests/log4cplus_c/run.sh
-rwxr-xr-x 1 root root 751 Aug 28 tests/msm_basic/run.sh
-rwxr-xr-x 1 root root 472 Aug 28 tests/res_man_dependency/run.sh
-rwxr-xr-x 1 root root 465 Aug 28 tests/res_man_ipc/run.sh
-rwxr-xr-x 1 root root 789 Aug 28 tests/res_man_multi_process/run.sh
-rwxr-xr-x 1 root root 469 Aug 28 tests/res_man_private_client/run.sh
-rwxr-xr-x 1 root root 492 Aug 28 tests/res_man_public_client/run.sh
-rwxr-xr-x 1 root root 311 Aug 28 tests/virt_mem_config/run.sh
lrwxrwxrwx 1 root root 6 Aug 28 var/run]
我尝试的技巧是添加一个空格,这在我的输入文件中有保证,这可以在控制台中使用,但不能在分配给变量时使用。
grep " tests/aes/run.sh" /home/rajesh/rootfs.layout
脚本中的行
pcline=`grep \"" $file1"\" $file2` # grep for file1 in file2 contents
如果我在此脚本中犯了任何错误,请告诉我。
答案 0 :(得分:1)
您可以像这样使用egrep
:
egrep "(^| )$file1( |$)" "$file2"
如果file1="run"
,那么上面的命令将匹配字符串run
,前面是行开头或空格,后跟空格或行尾。